0=-16t^2+140+4

Solution for 0=-16t^2+140+4 equation:

0=-16t^2+140+4

We move all terms to the left:

0-(-16t^2+140+4)=0

We add all the numbers together, and all the variables

-(-16t^2+140+4)=0

We get rid of parentheses

16t^2-140-4=0

We add all the numbers together, and all the variables

16t^2-144=0

a = 16; b = 0; c = -144;
Δ = b2-4ac
Δ = 02-4·16·(-144)
Δ = 9216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9216}=96$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-96}{2*16}=\frac{-96}{32}
=-3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+96}{2*16}=\frac{96}{32}
=3
$